POJ 1003 Hangover - 追着光梦游

POJ 1003 Hangover

ACM解题报告 2017-08-23

Hangover

Description
How far can you make a stack of cards overhang a table? If you have one card, you can create a maximum overhang of half a card length. (We're assuming that the cards must be perpendicular to the table.) With two cards you can make the top card overhang the bottom one by half a card length, and the bottom one overhang the table by a third of a card length, for a total maximum overhang of 1/2 + 1/3 = 5/6 card lengths. In general you can make n cards overhang by 1/2 + 1/3 + 1/4 + ... + 1/(n + 1) card lengths, where the top card overhangs the second by 1/2, the second overhangs tha third by 1/3, the third overhangs the fourth by 1/4, etc., and the bottom card overhangs the table by 1/(n + 1). This is illustrated in the figure below.


Input
The input consists of one or more test cases, followed by a line containing the number 0.00 that signals the end of the input. Each test case is a single line containing a positive floating-point number c whose value is at least 0.01 and at most 5.20; c will contain exactly three digits.
Output
For each test case, output the minimum number of cards necessary to achieve an overhang of at least c card lengths. Use the exact output format shown in the examples.
Sample Input
1.00
3.71
0.04
5.19
0.00
Sample Output
3 card(s)
61 card(s)
1 card(s)
题目大意
虽然题目还给出了图片,以及很多描述,但是我们可以总结为:
输入一个浮点数,记为c,找出累加式1/2 + 1/3 + 1/4 + ... + 1/(n + 1)大于等于c的最小的n,0.01<=c<=5.20,输入0.00代表输入结束。

#include<stdio.h>
int main()
{
    double res[300]={0.0,0.5},c=0.0;
    int i;
    for(i=2;i<300;i++)
    res[i]=res[i-1]+1.0/(i+1);
    //scanf("Lf",&c);//scanf double 输入总是0 
    //cout<<c<<endl;
    while(scanf("%lf",&c)&&c)
{
        for(i=1;i<300;i++)
    {
        if(res[i]>=c)
        {printf("%d card(s)\n",i);
        break;
        }
    }
}
    return 0;
} 

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